$w_1=81[\cos(\dfrac{\pi}{3})+i\sin(\dfrac{\pi}{3})]$ $w_2=9[\cos(\dfrac{\pi}{6})+i\sin(\dfrac{\pi}{6})]$ Express the quotient, $\dfrac{w_1}{w_2}$, in polar form. The angle should be given in radians. $\dfrac{w_1}{w_2}= $
Answer: Background For any two complex numbers $z_1$ and $z_2$ (whose radii are $r_1$ and $r_2$ and angles are $\theta_1$ and $\theta_2$ ): The radius of $\dfrac{z_1}{z_2}$ is the quotient of the original radii, $\dfrac{r_1}{r_2}$. The angle of $\dfrac{z_1}{z_2}$ is the difference of the original angles, $\theta_1 - \theta_2$. In other words, suppose the polar forms of $z_1$ and $z_2$ are as follows, $z_1 = r_1[\cos(\theta_1) + {i}\sin(\theta_1)]$ $z_2 = r_2[\cos(\theta_2) + {i}\sin(\theta_2)]$, then the polar form of their quotient is: $\dfrac{z_1}{z_2} = \dfrac{r_1}{r_2}[\cos(\theta_1 - \theta_2) + {i}\sin(\theta_1 - \theta_2)]$. [How do we get this?] Finding the radius of $\dfrac{w_1}{w_2}$ $w_1=81[\cos(\dfrac{\pi}{3})+i\sin(\dfrac{\pi}{3})]$ $w_2=9[\cos(\dfrac{\pi}{6})+i\sin(\dfrac{\pi}{6})]$ Here, $r_1=81$ and $r_2=9$. Therefore, the radius of $\dfrac{w_1}{w_2}$ is $\dfrac{r_1}{r_2}=9$. Finding the angle of $\dfrac{w_1}{w_2}$ $w_1=81[\cos(\dfrac{\pi}{3})+i\sin(\dfrac{\pi}{3})]$ $w_2=9[\cos(\dfrac{\pi}{6})+i\sin(\dfrac{\pi}{6})]$ Here, $\theta_1=\dfrac{\pi}{3}$ and $\theta_2=\dfrac{\pi}{6}$. Therefore, the angle of $\dfrac{w_1}{w_2}$ is $\theta_1-\theta_2=\dfrac{\pi}{3}-\dfrac{\pi}{6}=\dfrac{\pi}{6}$ Summary We found that the radius of $\dfrac{w_1}{w_2}$ is $9$ and its angle is $\dfrac{\pi}{6}$. Therefore, $\dfrac{w_1}{w_2}=9\left(\cos\left(\dfrac{\pi}{6}\right)+i\sin\left(\dfrac{\pi}{6}\right)\right)$